HAL Id: hal-03015447
https://hal.archives-ouvertes.fr/hal-03015447
Submitted on 19 Nov 2020
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Analytical modeling and design of high temperature
superconducting machine
Thierry Lubin
To cite this version:
Thierry Lubin. Analytical modeling and design of high temperature superconducting machine. École thématique. HTS Motors School, Nancy, France. 2020. �hal-03015447�
1
A
NALYTICAL MODELING AND DESIGN
OF HTS MACHINES
High Temperature Superconducting Motor School
August 31 – September 4, 2020, Nancy, France
Thierry Lubin
Research Group in Electrical Engineering (GREEN), Nancy, France Email: thierry,lubin@univ-lorraine,fr
2
❑ Speaker Biography (short)
❑ Topology of the studied HTS machine
❑ 2D (Semi)-Analytical Model : assumptions, equations,
solution…and discussion
❑ Back EMF and Torque computation
❑ Results: Matlab Programm, comparison with FEM
❑ Multi-Objective Design Optimization with GA:
P = 3 MW, N = 3000 rpm, PTM > 20 kW/Kg
(Electrical aircraft application, or other…)
Discussions and questions during the presentation
3
Research topics:
2003: Ph.D. degree in electrical engineering, Nancy, France.
“Modeling and control of a Synchronous Reluctance Machine with magnetic saturation effects”
2016: HDR (Habilitation) degree in electrical engineering, Nancy, France (confers the right to be sole advisor of a PhD student).
“Contribution to the analytical modeling of electromagnetic
actuators”
1994: Msc. in electrical engineering, Paris 6, France + “agrégation” in applied physical sciences (to teach in high school),
Since 2007 : Associate Professor in electrical engineering at Nancy Before 2007: High school teacher
2D and 3D analytical modeling of electrical machines for their design Contactless torque transmissions by magnetic gears and couplers Applied superconductivity in electrical engineering
Teaching topics (Master degree):
Electrical machines: modeling, design and control.
AC electrical power network: modeling, power quality, protection.
High power wind turbine: electrical part (machine, power electronics, control).
Author and co-author of more than 100 research papers in peer-reviewed journals and international conferences :https://cv.archives-ouvertes.fr/thierry-lubin
4
Why studied HTS machines?
Off-Shore Wind turbine
EcoSwing Project (2019)
3.6MW ; 15 rpm
(- 40% weight compare to
conventional system)
More Electrical Aircraft
Motor propulsion
1-5 MW ; 3000-6000 rpm
Power to Mass ratio > 20 kW/kg
Impossible with conventional
technology (5 kW/Kg, PMs, copper)
For high power electrical machines (MW)
5
Topology of the studied HTS machine
Superconducting Synchronous Radial flux machine (axial flux is possible but more complicated to model due to the 3D effect….)
Slotless machine: inductor with HTS winding (not possible with copper due to the Joule losses, sometimes with PM…)
Rotor HTS winding or bulk: ±Jc with Jc = 200 A/mm2at 20-30K (more complex
model next days, the cooling is not studied here)
Three phase stator AC winding: Jsrms = 20A/mm2(possible with water or oil forced
cooling, look at electrical vehicle for example, Litz wire to reduce AC losses)
Stator coils: 3 phases AC winding (copper) Stator iron yoke (or fiber…)
Air-gap, cryostat HTS winding : DC
with ±Jc
Rotor iron yoke (or fiber…)
6
HTS wires : J
c
(B,T) characteristics
6
Jc = 200 A/mm
2at 30K is correct but this should be include in the design
procedure (not done here)… Perhaps the maximal constraint is on the end-winding
(3D model…..)
7
What we want to obtain « by hand »!
No-load: back EMF computation Full-load (90°): Torque computation
A very fast and accurate model is needed if we want to use it with
an optimization procedure : Analytical model
First step design (cooling, ac losses….)!
We will consider all the source terms (rotor and stator) in the analytical model but this is not mandatory, it depends on what we want to study!
- The torque can be computed by the Laplace force by only knowing the rotor magnetic field distribution.
- But if we want to know the armature reaction (harmonic terms) on the HTS winding (AC losses computation…), we have to compute it!
8
Physical and Geometrical parameters
p : pole-pairs number
L : axial length
R1: inner radius of the rotor a = R2-R1 : thickness rotor yoke b = R3-R2 : thickness HTS winding c = R4-R3 : air-gap thickness
d = R5-R4 : thickness stator winding e = R6-R5 : 2nd air-gap thickness
f = R7-R6: thickness stator yoke α : outer opening of HTS coils β : innerr opening of the HTS coils : opening of the stator coils
0<;;;<1
s: stator yoke relative permeability r: rotor yoke relative permeability Jc : critical current of the HTS coils Krfill: fill factor HTS coils
Jc = 200 A/mm2 ; K
rfill = 0,7
Jsrms: stator RMS current density Ksfill: fill factor stator coils
Jsrms = 20 A/mm2 ; K
sfill = 0,5 3-phase distibuted stator winding
9
Model and Assumptions
8 homogeneous cylindrical regions with periodicity of 2
/p
-
Perfect cylindrical geometry: cylindrical coordinates (r,
)
-
End-effects are neglected: 2D analytical model
-
Constant permeability for the iron parts: magnetic saturation is not
taken into account (be careful that B<2T in the iron for the simulation!!)
-
Current in the coils is imposed (stator and rotor): equivalent current
10
2D field problem in cylindrical coordinates
Current density J = (0, 0, J
z
)
Vector magnetic potential A = (0, 0, A
z
)
Magnetic field strength H = (H
r
, H
, 0)
Flux density B = (B
r
, B
, 0)
11
Maxwell’s equations and Boundary
conditions at radius R
i
0
=
=
H
J
B
with
(linear and no magnet)
current density in the coils
0
:
=
rB
H
J
Magnetic vector potential formulation :
with
0
0
=
B
→
B
=
A
=
A
( )
,
zA r
=
zA
u
(regions 3 and 5)
(for the other regions)
0
0
= −
=
A
J
A
Ri i i+1 Continuity of the normal component of the flux density:
Continuity of the tangential component of the magnetic
field:
(
,
)
1(
,
)
(
,
)
1(
,
)
ri i ri i i i i iB
R
=
B
+R
→
A R
=
A
+R
(
)
(
)
1 1 11
1
,
,
i i i i i i i i R R i i
A
A
H
R
H
R
r
r
+
+ +
=
→
=
12
Partial Differential Equations to solve
( )
( )
2 2 3 3 3 0 2 2 2 2 2 5 5 5 0 2 2 2 2 2 2 2 21
1
for region 3 (HTS windings)
1
1
for region 5 (stator windings)
1
1
0 for other regions (i = 1, 2, 4, 6, 7 an
r s i i i
A
A
A
J
r
r
r
r
A
A
A
J
r
r
r
r
A
A
A
r
r
r
r
+
+
= −
+
+
= −
+
+
=
d 8)
Source terms: Stator and rotor current density distribution
Periodic problem with a period equal to 2/p, the solution in each
region can be write as
( )
(
(
)
)
1,3,5...sin
N r rn nJ
J
np
==
+
( )
( )
1,3,5...sin
N s sn nJ
J
np
==
With Jrn and Jsn to be determinate (next slides), is the load angle attached to the rotor
( )
( )
(
(
)
)
1,3,5...,
sin
N i in nA r
A
r
np
==
+
13
Complex notation (more compact)
( )
( )
(
(
)
)
( )
( )
1,3,5... 1,3,5...,
sin
,
Im
N N jnp in i in i n nA r
A r
np
A r
A
r e
= =
=
+
→
=
( )
( )
2 2 2 2 2 3 3 2 3 3 3 3 rn 0 0 2 2 2 21
1
n n
-
nr J
rA
A
A
d A
d A
J
r
r
np
A
r
r
r
r
dr
dr
+
+
= −
→
+
= −
The PDE (region 3) becomes an ODE complex equation (Euler):
Solution of the Euler equation:
( )
( )
3 1 2
np np
n p
A
r
=
K r
+
K r
−+
A
r
Where K1 and K2 are the integration constants that will be computed from the interface conditions and Ap(r) is the particular solution :
( )
( )
(
)
2 0 2 2 0for np
2
4
1
1 4 ln
for np
2
16
rn p rnr
J
np
A
r
r
r J
−
=
−
=
14
Solution in the different regions
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
1 1 2 2 3 3 4 5 3 4 6 7 5 8 9 5 6 10 11 7 12 13 8 14 np n np np n np np n p np np n np np n p np np n np np n np nA
r
K r
A
r
K r
K r
A
r
K r
K r
A
r
A
r
K r
K r
A
r
K r
K r
A
r
A
r
K r
K r
A
r
K r
K r
A
r
K r
− − − − − − −=
=
+
=
+
+
=
+
=
+
+
=
+
=
+
=
( )
( )
(
)
2 0 2 3 2 0for np
2
4
1
1 4 ln
for np
2
16
rn p rnr
J
np
A
r
r
r J
−
=
−
=
( )
( )
(
)
2 0 2 5 2 0for np
2
4
1
1 4 ln
for np
2
16
sn p snr
J
np
A
r
r
r J
−
=
−
=
We have 14 coefficients to determine (K
1to K
14) from the interface
conditions; Expression of J
rnand J
snto be determinate
15
Relations between the unknown
coefficients (1/3)
Ri i i+1( )
( 1)( )
( 1) 11
1
i i i i in i n i n in R i i RA
R
A
R
A
A
r
r
+ + +
=
=
It’s up to you!
Determine the 14 relations between the coefficients
and write them in matrix form.
Relations at the interface between 2 regions:
We have 7 interfaces (R
1to R
7) with 2 relations by interface that corresponds
16
Relations between the unknown
coefficients (2/3)
(
)
( )
(
)
( )
(
)
( )
1 1 2 1 3 1 1 1 1 1 1 1 2 1 3 1 2 2 2 2 3 2 4 2 5 2 0 2 2 1 1 1 1 2 2 2 3 2 4 2 5 2 0 2 2 3 4 3 5 3 0 2 21
4
1
2
4
np np np np np np r np np np np rn np np np np rn r np npK R
K R
K R
r
R
K R
K R
K R
R
K R
K R
K R
K R
J
np
r
R
R
K R
K R
K R
K R
J
np np
R
K R
K R
np
r
R
− − − − − − − − − − − − − −
=
+
=
→
=
−
+
=
+
+
−
=
→
−
=
−
+
−
+
+
=
→
( )
(
)
( )
( )
(
)
6 3 7 3 1 1 3 1 1 4 3 5 3 0 2 6 3 7 3 2 4 6 4 7 4 8 4 9 4 0 2 4 1 1 1 1 4 6 4 7 4 8 4 9 4 0 24
2
4
4
2
4
np np rn np np np np rn np np np np sn np np np np snJ
K R
K R
R
K R
K R
J
K R
K R
np np
R
K R
K R
K R
K R
J
np
r
R
R
K R
K R
K R
K R
J
np np
− − − − − − − − − − − − − − −
=
+
−
−
+
=
−
−
+
=
+
+
−
=
→
−
=
−
+
−
17
Relations between the unknown
coefficients (3/3)
( )
( )
(
)
(
)
2 5 8 5 9 5 0 2 10 5 11 5 5 1 1 5 1 1 8 5 9 5 0 2 10 5 11 5 10 6 11 6 12 6 13 6 6 1 1 1 1 10 6 11 6 12 6 13 64
2
4
1
np np np np sn np np np np sn np np np np np np np np sR
K R
K R
J
K R
K R
np
r
R
R
K R
K R
J
K R
K R
np np
K R
K R
K R
K R
r
R
K R
K R
K R
K R
r
− − − − − − − − − − − − − − − −
+
+
=
+
−
=
→
−
+
=
−
−
+
=
+
=
→
−
=
−
(
)
12 7 13 7 14 6 7 1 1 1 12 7 13 7 14 71
np np np np np np sK R
K R
K R
R
K R
K R
K R
− − − − − − −
+
=
=
→
−
= −
Valid for np 2. For np = 2 (that corresponds only to the case n = 1 and p = 2), see the previous expression for the particular solution (slide 13).
In the previous relations, s and r are the relative permeability, s = r = 1 if regions 2 and 7 are made with no-magnetic material, that corresponds to a totally ironless machine!
18 11 11 11 21 21 21 32 33 34 35 42 43 44 45 54 55 56 57 64 65 66 67 76 77 78 79 86 87 88 89 98 99 910 911
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
1 2 3 4 5 6 7 8 9 10 108 109 1010 1011 1110 1111 1112 1113 12 1211 1212 1213 1312 1313 1314 1412 1413 14140
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
K
K
K
K
K
K
K
K
K
M
M
M
M
K
M
M
M
M
M
M
M
M
M
M
M
M
M
M
3 4 5 6 7 8 9 10 11 12 13 140
0
0
0
0
0
S
S
S
S
S
S
S
S
K
K
K
K
=
Writing in Matrix form (Global matrix)
Where M11, M12 , …. are sub-matrix (n-by-n matrix, diagonal, where n is number of harmonics considered )
S3, S4….. are the source terms (vectors of length n)
19
Sub-Matrix form
( )
2 2 2 3 4 5 2 2 2 2 0 24
np np np np rnR
R K
R
K
R K
R
K
J
np
− −+
−
−
=
−
M
31
M
32
M
33
M
34( )
3S
M
11
M
12
M
13 1 2 3 1 1 10
np np npR K
−
R K
−
R
−K
=
Important note : ill-conditioned global matrix and numerical problem
First line of the global matrix:
Third line of the global matrix:
For 3 harmonics (n = 1, 3, 5) and p = 4
4 1 12 11 1 20 10
0
0
0
0
0
R
M
R
R
=
4 1 12 13 1 20 10
0
0
0
0
0
R
M
R
R
− − −
= −
( )
2 0 2 1 2 0 2 3 3 2 0 2 512
140
396
r r rR
J
R
S
J
R
J
=
For example, if we consider R1 = 0,2 m for 3 harmonics (n = 1, 3, 5) and p = 4
3 9 11 141.10
0
0
0
4.10
0
0
0
1.10
M
− − −
=
8 13 13625
0
0
0
2.10
0
0
0
9.10
M
= −
Very large numbers and very small numbers in the topological matrix: numerical problem of precision during the inversion: it is better to write the solution as dimensionless coefficients (radius ratio), but this is not dealt with here…….You can do it!
20
Matrix form : extract of the Matlab
21
Source terms : HTS current density J
r
(
)
p : pole-pairs number
α : outer opening of HTS coils β : inner opening of the HTS coils : rotor position (load torque) Jc : critical current of the HTS coils Krfill: fill factor for the HTS coils
Jc = 300 A/mm2 ; K rfill = 0,7
( )
(
(
)
)
1,3,5...sin
N r rn nJ
J
np
==
−
4
cos
cos
2
2
rfill c rnK
J
J
n
n
n
=
−
In complex notation:
J
rn=
J e
rn −jnp22
Source terms : HTS current density J
r
(
)
4
cos
cos
2
2
rfill c rnK
J
J
n
n
n
=
−
p = 2;α = 0.9; β =0.5 Jc = 300 A/mm2; K rfill = 0,723
Source terms: three-phase AC winding J
s
(
)
3-phase distibuted stator winding
p : pole-pairs number
: opening of the stator coils Jsrms : stator RMS current density Ksfill: fill factor stator coils
Jsrms= 20 A/mm2 ; K
sfill = 0,5
The 3 phases windings are fed by AC
3 phases current
→ Rotating field
( )
( )
( )
(
)
( )
(
)
2 cos
2 cos
2
3
2 cos
2
3
A sfill s rms B sfill s rms B sfill s rmsJ
t
K
J
t
J
t
K
J
t
J
t
K
J
t
=
=
−
=
+
24
Source terms: three-phase AC winding J
s
(
)
( )
( )
( )
(
)
( )
(
)
2 cos
2 cos
2
3
2 cos
2
3
A sfill s rms B sfill s rms B sfill s rmsJ
t
K
J
t
J
t
K
J
t
J
t
K
J
t
=
=
−
=
+
(
)
(
)
(
)
0
2
1
0
2
2
1
0
2
2
A sfill s rms B sfill s rms B sfill s rmsJ
t
K
J
J
t
K
J
J
t
K
J
= =
= = −
= = −
At t = 0
Field distribution at t = 0Current density distribution at t = 0
( )
( )
1,3,5...sin
N s sn nJ
J
np
==
4
2 sin
sin
1 cos
2
2
6
3
sn sfill srmsJ
K
J
n
n
n
n
=
−
with25
Source terms: three-phase AC winding J
s
(
)
p = 2; ν = 0.9
Jsrms= 20 A/mm2; K
rfill = 0,5
4 2
2 sin sin 1 cos
2 6 3 sn sfill srms J K J n n n n
= − 26
More complicated 3-phase AC winding to
reduce the harmonic level
Reduce the harmonic content of the stator winding could be good to
avoid AC losses in the HTS coils (due to parasitic rotating field)
Distributed winding and chording (short pitched coil)
Qs : number of stator “slots” p: number of pole pairs
q: number of slots per pole and per phase
6
s
Q
q
p
27
More complicated 3-phase AC winding to
reduce the harmonic level
( ) ( )
4
2
2 sin
sin
1 cos
2
6
3
sn sfill srms d pJ
K
J
n
n
n
K
n K
n
n
=
−
( )
( )
sin
is the distribution factor ;
sin(
) is the pitch factor
2
sin
s d p snpq
Q
OUV
K
n
K
n
n
D
q
np
Q
=
=
28
Expression of the Flux density
( )
( )
(
(
)
)
( )
( )
1,3,5... 1,3,5...,
sin
,
Im
N N jnp in i in i n nA r
A r
np
A r
A
r e
= =
=
+
→
=
The magnetic vector potential in each region is written as
( )
with
A r
i,
=
=
i i i zB
A
A
u
In cylindrical coordinates, we have:
1
i ii
A
A
B
r
r
=
−
u
r
u
( )
( )
( )
( )
1,3,5... 1,3,5...Radial component of
,
Im
Tangential component of
,
Im
N jnp rin ri n N jnp in i nB
r
B
r e
B
r
B
r e
= =
→
=
→
=
B
B
( )
( )
( )
( )
1
rin in in inB
r
jnp A
r
r
d A
r
B
r
dr
=
= −
29
Results and validation with FE simulations
p = 2 : pole-pairs number L= 30 cm : axial length
R1 = 3 cm: inner radius of the rotor a = 2 cm : thickness rotor yoke b = 1 cm : thickness HTS winding c = 0.5 cm : air-gap thickness
d = 1.5 cm : thickness stator winding e = 0.5 cm : 2nd air-gap thickness
f = 1.5 cm : thickness stator yoke α = 0,777 : outer opening of HTS coils β = 0.333 : inner opening of the HTS coils = 0.666 : opening of the stator coils s = 1000 : stator relative permeability r= 1000 : rotor relative permeability
Jc = 200 A/mm2 : critical current of the
HTS coils (to avoid magnetic saturation) Krfill = 0.7 : fill factor for the HTS coils Jsrms = 20 A/mm2 : stator RMS current
density
30
31
Flux density in the middle of the air-gap
under load condition (
= 90°)
Flux line under Full-load condition (FEMM software)
Torque with FEM: T = 505 Nm in 1s
Torque with Analytical: T = 504 Nm in 20ms Computational time 50 times better: good for
32
Torque computation: Maxwell Stress Tensor
The torque is compute in the middle of the air-gap (region 4) : less harmonic terms are needed(
) (
)
2 2 4 4 0 0,
,
m r m t mLR
T
B
R
B
R
d
=
3 42
mR
R
R
=
+
(
) (
)
( ) ( )
2 4 4 4 4 0,
,
r r m m m mB
R
B
R
d
B
R
B
R
=
With the complex notation, we obtain:
( )
(
)
( )
(
)
4 6 7 1 1 4 6 71
np np r m m m m np np m m mB
R
jnp
K R
K R
R
B
R
np K R
K R
− − − −=
+
= −
−
See slide 27; For harmonic np we have :
( )
(
)
2
6 7 6 7
0 1,3,5...
where is the real part
n
L
T
j np
K K
K K
==
−
33
Static and dynamic torque for the studied
example
Static torque
Dynamic torque
We change the rotor position: 0 < < 90° The current in the stator is fixed: (I ; -I/2; -I/2)
Incremental variation of the rotor position from the maximal value of the torque: 90°+
The current in the stator coils change as well (current control with inverter and encoder):
( )
(
)
( )
(
)
( )
(
)
2 cos 2 cos 2 3 2 cos 2 3 A sfill s rms B sfill s rms B sfill s rms J K J p J K J p J K J p = = − = + Red: 10 harmonicsBlack : fundamental only Red: 10 harmonics
34
Flux and Back-EMF expression
Magnetic flux is obtained from the mean value of magnetic vector
potential in a stator coil cross-section (Green-Ostrogradsky theorem)
( )
5 4 6 5 6,
R p slot slot R pL
A r
rdrd
S
− − =
( )
5 4 6 5 62
,
R p phase slot R ppL
A r
rdrd
S
− −
=
Regular distributed winding (*2p)
( )
( )
(
)
2 2 2 2 5 4 5 4 8 9 1,3,5... 4 4 5 8 5 4 9 44
1
sin
Im
for
2
6
2
2
2
1
sin
Im
ln
f
3
4
np np np np phase slot n phase slotR
R
R
R
pL
np
j K
K
np
S
np
p
np
np
R
pL
j
K
R
R
K
S
p
R
+ + − + − + =
−
−
=
−
+
−
=
−
+
or
np =
2
The back EMF is obtained from the derivative of the flux
( )
phasewith
( is the mechanical position)
a c
d
d
E
N
d
dt
= −
=
35
Flux and Back-EMF computation for the
studied example (for 1 conductor/slot)
No load (J
srms= 0 A/mm
2) ;
= 314rad/s
Red: 10 harmonics ; Black : fundamental only
Rapid verification:
4 2 6 4
max
3 with 504 and 314 /
We have 3.7923.10 and 0.5 20.10 3.7923.10 3792 13.9 19.8 (for 1 conductor per slot!)
3
slot sfill srms slot
EI C C Nm rd s S m I k J S A C E A E V I − − = = = = = = = = = → =
36
Design Optimization of an HTS machine
More Electrical Aircraft
Motor propulsion
P= 3 MW ; N = 3000 rpm
→ T = 9500 Nm
Power to Mass ratio > 20 kW/kg ????
Genetic Algorithm available in Matlab Optimization Toolbox
A genetic algorithm (GA) is a method for solving both constrained and unconstrained optimization problems based on a natural selection process that mimics biological evolution. The algorithm repeatedly modifies a population of individual solutions. At
each step, the genetic algorithm randomly selects individuals from the current population and uses them as parents to produce the children for the next generation.
37
Mono-objective Design Optimization
Constraint
→ T = 9500 Nm
1 objective
→ Minimum active mass (HTS and copper coils
Totally iron less machine (µ
s= µ
r=1) ; R
1= R
2; R
5= R
6= R
7p = x(1) : pole-pairs number
R1 = x(2) : inner radius of the rotor a = x(3)= 0 cm : thickness rotor yoke b = x(4) : thickness HTS winding c = 1 cm : air-gap thickness (fixed) d = x(5) : thickness stator winding e = x(6) = 0 cm : 2nd air-gap thickness
f = x(7) = 0 cm : thickness stator yoke L= x(8) : axial length
α = x(9) : outer opening of HTS coils β = x(10) : inner opening of the HTS coils = x(11) : opening of the stator coils s = 1 : stator relative permeability r = 1 : rotor relative permeability Jc= 200 and 300 A/mm2 ; K
rfill = 0.7
Jsrms= 20 A/mm2 ;K
38
Active mass expression
We have to write the active mass (HTS and copper coils) as a function of the
geometrical parameters
(
2 2)
5 4
Volume active parts :
V
apcop=
R
−
R k
sfillL
Copper coils
cop 8960kgm 3:
−
=
(
)(
)
3 2 2
5 4 5 4
Volume End winding part :
4
sfill EWcopR
R
R
R k
V
p
−
+
=
(
)
Copper mass :
M
cop=
copV
apcop+
V
EWcop(
)
(
2 2)
3 2Volume active parts :
V
apHTS=
−
R
−
R k
rfillL
HTS coils
HTS =5500kgm−3:
(
)(
)
(
)(
)
3 2 2
3 2 3 2
Volume End winding part :
8
rfill EWHTSR
R
R
R k
V
p
+
−
−
+
=
(
)
HTS mass :
M
HTS=
HTSV
apHTS+
V
EWHTS39
Design Optimization with Matlab Toolbox
We have to define two Matlab function functions: constraint.m and objective.m
constraint.m: return the value of the constraint (torque)
Objective.m: return the value of the objective (Mass of the windings)
We have to choose an optimization method: mono objective Genetic Algorithm
(
ga.m
)
We have 11 variables with 1 integer (number of pole-pairs). We have to choose
lower and upper bounds for this variables (not easy, trial and errors….)
Variable Number Lower Upper
p Pole-pairs number (integer) x(1) 1 10
R1 Inner radius of the rotor (cm) x(2) 10 40
a Thickness rotor yoke (cm) x(3) 0 0
b Thickness HTS winding (cm) x(4) 0.5 2
d Thickness stator winding (cm) x(5) 0.5 5
e 2ndair-gap thickness (cm) x(6) 0 0
f Thickness stator yoke (cm) x(7) 0 0
L Axial length (cm) x(8) 20 50
α Outer opening HTS coils x(9) 0 1
Inner opening HTS coils x(10) 0.1 1
40
Design Optimization with Matlab Toolbox
41
Design Optimization with Matlab Toolbox
42
Design Optimization with Matlab Toolbox
To reduce the mass, the algorithm tends to maximize the radius and the pole
43
Design Optimization with Matlab Toolbox
44
Design Optimization with Matlab Toolbox
If we have a constraint on the radius, we can impose R1 = 25cm (for example)
Variable Number Lower Upper
p Pole-pairs number (integer) x(1) 1 10
R1 Inner radius of the rotor (cm) x(2) 25 25
a Thickness rotor yoke (cm) x(3) 0 0
b Thickness HTS winding (cm) x(4) 0.5 2
d Thickness stator winding (cm) x(5) 0.5 5
e 2nd air-gap thickness (cm) x(6) 0 0
f Thickness stator yoke (cm) x(7) 0 0
L Axial length (cm) x(8) 20 50
α Outer opening HTS coils x(9) 0 1
Inner opening HTS coils x(10) 0.1 1
45
Design Optimization with Matlab Toolbox
46
Design Optimization with Matlab Toolbox
47
Design Optimization with Matlab Toolbox
48
Design Optimization with Matlab Toolbox
49
HTS winding verification for the critical
current density (needs to be done….)
This should be include in the design procedure…… Perhaps the maximal
50
HTS winding verification for the critical
current density (needs to be done….)
B on HTS wire is never greater than 1.65T; Under 30K Jc = 300A/mm2 can be
51
Multi-objective Design Optimization
gamultiobj.m
51
We have to define two Matlab function functions: constraint.m and objective.m
constraint.m: return the value of the constraint (torque)
Objective.m: return the value of the objectives
We have to choose an optimization method: multi objective Genetic Algorithm
(
gamultiobj.m
)
Variable Number Lower Upper
p Pole-pairs number (integer) x(1) 1 10
R1 Inner radius of the rotor (cm) x(2) 10 40
a Thickness rotor yoke (cm) x(3) 0 0
b Thickness HTS winding (cm) x(4) 0.5 2
d Thickness stator winding (cm) x(5) 0.5 5
e 2ndair-gap thickness (cm) x(6) 0 0
f Thickness stator yoke (cm) x(7) 0 0
L Axial length (cm) x(8) 20 50
α Outer opening HTS coils x(9) 0 1
Inner opening HTS coils x(10) 0.1 1
Opening of the stator coils x(11) 0.5 1
52
Multi-objective Design Optimization
gamultiobj
doesn’t manage the integer
variable!!!
53
Multi-objective Design Optimization:
Pareto front (J
c
= 300 A/mm2)
A
54
Multi-objective Design Optimization:
Results A and B
Variable Number Lower Upper A B
p Pole-pairs number (integer) 1 10 8 4
R1 Inner radius of the rotor (cm) 10 40 30.1 15.8
a Thickness rotor yoke (cm) 0 0 0 0
b Thickness HTS winding (cm) 0.5 2 1.2 2.1
d Thickness stator winding (cm) 0.5 5 1.17 1.84
e 2nd air-gap thickness (cm) 0 0 0 0
f Thickness stator yoke (cm) 0 0 0 0
L Axial length (cm) 20 50 30.2 33.3
α Outer opening HTS coils 0 1 0.99 0.99
Inner opening HTS coils 0.1 1 0.413 0.427
Opening of the stator coils 0.5 1 0.652 0.747
Mass Activ mass, windings (kg) 50.6 59.7
Rext External radius (cm) 33.4 20.7
55
Multi-objective Design Optimization:
Pareto front (J
c
= 200 A/mm2)
A
B
J
c=300A/mm
2and R
2= 25cm → M = 52.5kg, PTM = 57 kW/kg
J
c=200A/mm
2and R
2= 25cm → M = 67.6kg, PTM = 44 kW/kg
56
Design Optimization with iron yoke
We have to limit the maximal value of the flux density in the stator and rotor yokes
around the knee point of the B(H) characteristic (we have the analytical expression
of the flux density in the yokes), if not the model will be not valid……..
Supplementary constraint : B
yoke< B
knee(around 1,6T)
Knee point
To do….. Further results are expected
57
